- #36
pallidin
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K^2 said:I do understand the dynamics of a water balloon in a zero force field. You don't seem to.
You seem to think that there can be a net force acting on balloon under zero net force.
Ok, now your getting it.
K^2 said:I do understand the dynamics of a water balloon in a zero force field. You don't seem to.
You seem to think that there can be a net force acting on balloon under zero net force.
So... Universal law of gravity does not describe reality?pallidin said:No, no! Vector cancellation, mathematically, does cancel the force, BUT, this is not what happens in reality for your described circumstance.
pallidin said:No, no! Vector cancellation, mathematically, does cancel the force, BUT, this is not what happens in reality for your described circumstance.
This would be true if the net force at the very centre of the hollow Earth were zero, but non-zero (upwards) at all other points inside the cavity. But it's not true. The net force is zero everywhere within the hollow cavity. The molecules of water can't tell the difference between multiple forces summing to zero and no force at all. (To all intents and purposes there is no difference.)pallidin said:Imagine that you are a water-filled balloon in the center of a hollow earth.
The water IN THAT BALLOON WILL BE PULLED away(slightly) from it's center and optimize on the spherical edge inside the envelope of the balloon.
Curl said:... the point is that since the Earth is denser into core, going deep inside will INCREASE the gravitational force on a person because you are CLOSER to the dense part of the Earth (iron and nickel I hear?)
...
DrGreg said:The molecules of water can't tell the difference between multiple forces summing to zero and no force at all. (To all intents and purposes there is no difference.)
Well, nuts! I made a sign error there.D H said:Lighter? Yes. Considerably lighter? No. The effect is *tiny*. Let's say your father's friend had to carry 50 pounds of equipment. Going 1/2 mile (~800m) underground would make that equipment feel about 1/10 of an ounce lighter.
Yes and no. It does go to zero at the center, but the graph of gravitational acceleration inside the Earth is a bit strange. Gravitational acceleration increases with depth down to 500 km or so. Gravitation then decreases down to a depth of 1600 km or so, where it starts increasing again. It continues to increase down to 2900 km or so below the surface, where g reaches it global maximum of about 10.48 m/s2.K^2 said:Keep in mind that this increase doesn't last for very long. It does start to drop after some depth, and goes to zero at the center.
cragar said:would my clock tick faster at the center of the Earth ,
and can my atomic clock tell the difference between no gravitational field , and a field cancellation of vectors to produce a zero field .
That's the bit that's new to me. Do you have a link to any place that derives it?Dickfore said:According to the formula:
[tex]
g_{00} = 1 + \frac{2 \, \varphi}{c^{2}}, \; |\varphi| \ll \frac{c^{2}}{2}
[/tex]
Skimming through my copy of Rindler seems to suggest that, for some GR coordinates (static or stationary, I think), then there is such a thing as relativistic potential given not approximately but exactly byDickfore said:Of course it's an approximation, there is no meaning to define a gravitational potential in General Relativity.
elfboy said:Imagine falling into the earth. As you fall there is a semiphere pulling you up.
pallidin said:With all due respect, that simply is not true at all.
K^2 said:Look up the Gauss' Theorem. You seriously aren't helping yourself by arguing something from pure ignorance and refusing to look up something that actually explains the physics behind it.
Hi K^2, in this thread https://www.physicsforums.com/showthread.php?t=458218 in the relativity forum I give the metrics for outside a shell, inside the material of a shell and inside a vacuum cavity enclosed by the shell. Basically it agrees with your conclusion here.K^2 said:There is a metric that describes gravity inside a spherical body. Using that metric, you'd be able to derive time dilation. I think it should be maximum at the center, but I'm not sure.
This seems reasonable. If we consider the Earth plus its atmosphere, then the gravitational force at the top of the atmosphere is less than the force at the bottom of the atmosphere because the density of the atmosphere is significantly less than the density of the Earth and the inverse square law dominates. Consider a sphere of radius r1 with constant density p1 enclosed by a shell extending from r1 to r2 with constant density p2. If have done the calculations correctly, then the force on a particle at radius x (where r1<=x<=r2) is equal to the force on a particle at r1 when p2/p1 = r1/x. If p2 < p1*r1/x then the force at x inside the shell is greater than at the surface (r2) of the shell.JDługosz said:Maybe this will help: Imagine only a dense iron core, with the mantle and crust missing. You are on a tower 4000 miles from the center. As you descend, you get heavier because you get closer to the dense core.
Add a fluffy crust a few feet thick. When you move from the "outside" to the "inside" of this outer sphere, you decrease your weight by the gravity contributed by this light fluffy shell. Are there conditions such that the latter effect is smaller than the former? sure; if the crust is light enough and/or the core dense enough.
Continental rocks literally float on the mantle. So I would in fact expect that effect to occur, but can't tell without further information which change dominates.
Note: horizontal scales are different for the two plots!Subductionzon said:
And the accompanying image of density as you go down to the center: